From 676bbc8b743265a633fcc00eb729a9f81af60ee3 Mon Sep 17 00:00:00 2001
From: yuanguangxin <274841922@qq.com>
Date: Fri, 26 Jun 2020 00:11:23 +0800
Subject: [PATCH] update

---
 Rocket.md                                     | 2 +-
 src/回溯法/q40_组合总和2/Solution.java | 6 ++++++
 2 files changed, 7 insertions(+), 1 deletion(-)

diff --git a/Rocket.md b/Rocket.md
index 806dac7..3e21e08 100644
--- a/Rocket.md
+++ b/Rocket.md
@@ -865,7 +865,7 @@ void quick_sort(int a[], int low, int high){
 ```
 
 5. 堆排序：将待排序序列构造成一个大顶堆，此时，整个序列的最大值就是堆顶的根节点。将其与末尾元素进行交换，此时末尾就为最大值。然后将剩余n-1个元素重新构造成一个堆，这样会得到n个元素的次小值。如此反复执行，便能得到一个有序序列了。
-```java
+```
 public class Test {
 
     public void sort(int[] arr) {
diff --git a/src/回溯法/q40_组合总和2/Solution.java b/src/回溯法/q40_组合总和2/Solution.java
index 4faf116..ca44cfe 100644
--- a/src/回溯法/q40_组合总和2/Solution.java
+++ b/src/回溯法/q40_组合总和2/Solution.java
@@ -30,9 +30,15 @@ class Solution {
             if (!set.contains(candidates[i]) && target >= candidates[i]) {
                 stack.push(candidates[i]);
                 helper(candidates, target - candidates[i], i + 1, stack, res);
+                System.out.println(stack + "before");
                 stack.pop();
+                System.out.println(stack + "after");
                 set.add(candidates[i]);
             }
         }
     }
+
+    public static void main(String[] args) {
+        new Solution().combinationSum2(new int[]{10, 1, 2, 7, 6, 1, 5}, 8);
+    }
 }